# How do you factor 9x^2 - 8y^2?

Jul 13, 2015

$9 {x}^{2} - 8 {y}^{2} = \left(3 x - 2 \sqrt{2} y\right) \left(3 x + 2 \sqrt{2} y\right)$

#### Explanation:

This is a difference of squares, but only with the help of irrational coefficients.

$9 {x}^{2} - 8 {y}^{2} = {\left(3 x\right)}^{2} - {\left(\sqrt{8} y\right)}^{2} = \left(3 x - \sqrt{8} y\right) \left(3 x + \sqrt{8} y\right)$

using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = 3 x$ and $b = \sqrt{8} y$

Note further that $\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2}$

So we can express the factorisation as

$9 {x}^{2} - 8 {y}^{2} = \left(3 x - 2 \sqrt{2} y\right) \left(3 x + 2 \sqrt{2} y\right)$